If I were to roll a die $y$-number of times, how do I calculate the probability of rolling a $6$ at least $x$-times?
Imagine that two people are making a bet over the outcome of a die roll. If the result is $1-5$, Tod pays Frank $100$ dollars. If the result is a $6$, Frank pays Tod $1000$ dollars. If this bet was repeated $60$ times, what is the probability that Tod will at least break even?
Use binomial distribution $Prob.=\sum_{k=x}^y\binom{y}{k}p^k(1-p)^{y-k}$ where $p=\frac{1}{6}$
To get break even: number you need, $1000n -100(60-n) \ge 0$ or $n=6$.
Prob. of break even : use above formula for $y=60$ and $x=6$.
For calculation, get $1-Prob.=\sum _{k=0}^5$